Thermodynamics - heat, internal energy, work

[Ampere]

Homework Statement

Heat flows into an ideal gas at a constant volume. The pressure increases from 1.5 atm to 5.5 atm. Next the gas is compressed at constant pressure from 5.0 L to 2.5 L and goes back to its original temperature.

1.What is the total work done on the gas in the process?
2.What is the total change in internal energy?
3.What is the total heat flow of the process?

Homework Equations

U = Q + W
W = -P*ΔV

The Attempt at a Solution

1. No work is done in the first process because the volume is constant. The work done on the gas in the second process is W = -(5.5 atm) * (2.5 L - 5.5 L) = 1390J

2. The change in internal energy is 0 because there is no net change in temperature.

3. 1390J of heat must have left the system because U = Q + W.

Is this right?

 

[TSny]

Hello, Ampere.

Your process looks right, but I don't agree with your conversion from L-atm to Joules.

[EDIT: It's not the conversion, it's a typo in your equation W = -(5.5 atm) * (2.5 L - 5.5 L) = 1390J
where you wrote 5.5 L instead of 5.0 L.

I agree with your answer now.]

 

[Ampere]

Yes, I meant (2.5 L - 5.0 L). Thanks for the confirmation!